Domain and Range of Rational Functions

You have a rational function and the range isn't obvious. The graph might cross its horizontal asymptote, or a hole might remove a value from the range. The fix is to find the domain first, then solve for x or check the asymptote with a test. The domain and range of rational functions follow two rules: denominator zeros are out, and horizontal asymptotes often limit the outputs.

Domain first. A rational function f(x) = P(x)/Q(x) is undefined where Q(x) = 0. Set the denominator equal to zero and solve. Those x-values are excluded. Write the domain as a union of intervals. For f(x) = 1/(x − 2), Q(x) = x − 2 gives x = 2. Domain: (−∞, 2) ∪ (2, ∞). For f(x) = (x² − 4)/(x − 2), Q(x) = x − 2 gives x = 2 again, but the numerator also has a factor (x − 2). That creates a hole, not a vertical asymptote. Domain still excludes 2. OpenStax Precalculus 2e, Section 5.6, covers this distinction.

A vertical asymptote occurs at an x-value where Q(x) = 0 but P(x) ≠ 0 after simplification. The function blows up to ±∞ there. A hole occurs where a common factor cancels, leaving a missing point. Both are excluded from the domain. The difference: the range loses the y-value of the hole, while the vertical asymptote restricts the x-value only.

For (x² − 4)/(x − 2), cancel (x − 2) to get f(x) = x + 2, x ≠ 2. The hole is at (2, 4). The range is all real numbers except 4. Paul's Online Math Notes, Algebra: Domain and Range examples, uses this exact case. For 1/(x − 2), no cancelation, so a vertical asymptote at x = 2. The range excludes 0 because the horizontal asymptote is y = 0 and the function never reaches it.

Two methods exist for finding the range of a rational function. Method one: solve y = f(x) for x and find which y-values give a real x. Method two: identify the horizontal asymptote and test whether the function crosses it. Both require checking holes.

For f(x) = (2x + 1)/(x − 3), degrees are equal, so the horizontal asymptote is y = 2 (ratio of leading coefficients). Solve for x: y(x − 3) = 2x + 1 → yx − 3y = 2x + 1 → yx − 2x = 3y + 1 → x = (3y + 1)/(y − 2). The denominator y − 2 = 0 gives y = 2, so y = 2 is excluded. Range: (−∞, 2) ∪ (2, ∞). OpenStax Precalculus 2e, Section 3.2, shows this solving method.

The rule 'range excludes the horizontal asymptote value' fails if the function crosses the asymptote. Check by setting f(x) equal to the asymptote value and solving for x. If a real x exists, the range includes that value.

Example: f(x) = x/(x² + 1). Horizontal asymptote is y = 0 (numerator degree < denominator degree). Set x/(x² + 1) = 0 → x = 0. So the function crosses at (0, 0). The range includes 0. Actual range is [−0.5, 0.5] from calculus, but the key point: do not blindly exclude the asymptote value. Paul's Online Math Notes warns about this.

Example 1: f(x) = 1/(x − 2). Denominator zero at x = 2. Vertical asymptote. No cancelation. Domain: (−∞, 2) ∪ (2, ∞). Horizontal asymptote y = 0. Does it cross? 1/(x − 2) = 0 has no solution. Range: (−∞, 0) ∪ (0, ∞).

Example 2: f(x) = (x² − 4)/(x − 2). Factor numerator: (x − 2)(x + 2). Cancel common factor (x − 2). Simplified: f(x) = x + 2, x ≠ 2. Hole at (2, 4). Domain: (−∞, 2) ∪ (2, ∞). Range: (−∞, 4) ∪ (4, ∞). No horizontal asymptote (linear).

Example 3: f(x) = (2x + 1)/(x − 3). Denominator zero at x = 3. No cancelation. Vertical asymptote at x = 3. Domain: (−∞, 3) ∪ (3, ∞). Horizontal asymptote y = 2. Solve (2x + 1)/(x − 3) = 2 → 2x + 1 = 2x − 6 → 1 = −6, false. No crossing. Range: (−∞, 2) ∪ (2, ∞).

Find the domain and range of each function. Check for holes and asymptote crossings.

1. f(x) = 3/(x + 5)
2. f(x) = (x² − 9)/(x − 3)
3. f(x) = (4x − 1)/(2x + 3)
4. f(x) = (x² + 1)/(x − 1)

Answers: 1) Domain: (−∞, −5) ∪ (−5, ∞). Range: (−∞, 0) ∪ (0, ∞). 2) Domain: (−∞, 3) ∪ (3, ∞). Range: (−∞, 6) ∪ (6, ∞). 3) Domain: (−∞, −1.5) ∪ (−1.5, ∞). Range: (−∞, 2) ∪ (2, ∞). 4) Domain: (−∞, 1) ∪ (1, ∞). Range: (−∞, −2] ∪ [2, ∞).